# Math1110 (Spring 2009) Prelim 3 - Solutions

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1 Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 1 Queston 1. (16 ponts) Short answer. Math1110 (Sprng 2009) Prelm 3 - Solutons x a 1 (a) (4 ponts) Please evaluate lm, where a and b are postve numbers. x 1 x b 1 sn(x) (b) (4 ponts) Please evaluate lm x π 1 cos(x). (c) (4 ponts) Please evaluate lm x x2 e x. (d) (4 ponts) Please wrte the sum (n 1) n n sgma notaton and fnd a closed form expresson for t. What s the value of the sum when n = 10? Soluton to Queston 1. (a) usng l Hoptal s rule, we get (b) lm x π (c) Usng l Hoptal s rule twce we get (d) x a 1 lm x 1 x b 1 = lm ax a 1 h 1 bx = a b 1 b. sn(x) 1 cos(x) = sn π 1 cos π = 0 1 ( 1) = 0. lm x x2 e x = lm x x 2 = lm e x x 3x = lm e x x 2e x (n 1) n = (7 + ) When n = 10, then the sum equals = 7 + = 7n + (n (n 1))/2. 10 (10 1) 2 = = 115.

2 Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 2 Queston 2. (20 ponts) True/False. Determne whether the followng statements are true or false, and crcle your response. Please gve a bref explanaton (n a complete sentence!). (a) (4 ponts) If a dfferentable functon f s defned at c, and c s a crtcal pont of f, then f has ether a local mnmum or a local maxmum at c. TRUE FALSE (b) (4 ponts) The functon f(x) = sn(cos(x)) + x3 + 7 e x s ntegrable on the nterval [a, b] for any real numbers a and b. TRUE FALSE (c) (4 ponts) Let f(x) be a functon, and P n (x) ts order n Maclaurn polynomal. Suppose that P n (0.1) = and E n (0.1) Then we know for certan that f(0.1), rounded to two decmal places, s TRUE FALSE (d) (4 ponts) The area of the regon bounded by sn x and the x-axs, between x = π and x = π s computed by π π sn x dx. TRUE FALSE (e) (4 ponts) To estmate the area below the functon y = x between x = 0 and x = 2, we may wrte a Remann sum ( ( ) 2 2 S n = + 4) 2 n n. Ths sum S n s less than the actual area for all n. TRUE FALSE Soluton to Queston 2. (a) FALSE. The functon y = x 3 has a crtcal pont at x = 0, but nether a local max nor a local mn there. (b) TRUE. Ths functon s contnuous at all real numbers x : each of the functons sn(x), cos(x), x 3 + 7, e x s contnuous, and composton, sums, and quotents are contnuous (snce e x 0, there are no ssues when we have t n the denomnator). A theorem from 5.3 says that f f(x) s contnuous on [a, b], t s ntegrable on [a, b]. (c) FALSE. If E n (0.1) = 0.005, then f(0.1) = P n (0.1) + E n (0.1) = = whch s 3.72 when rounded to two decmal places. (d) FALSE. Because from π to 0, n the above ntegral, the area between π to 0 has a mnus sgn (below the x-axs. (The ntegral becomes actually 0 because of the symmetry of sn x around the orgn). (e) TRUE. The functon s decreasng on [0, 2] but stays postve.

3 Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 3 Queston 3. (20 ponts) Consder the functon f(x) = 1 x2 x 2 4. We wll get you started by computng the frst two dervatves: f (x) = (a) Fnd all crtcal and sngular ponts of f. (b) Fnd all asymptotes of f. (c) Where s f ncreasng and decreasng? (d) Where s f concave up and concave down? 6x (x 2 4) 2 and f (x) = 18x (x 2 4) 3. (e) Usng what you have calculated n the frst four parts of ths problem, please graph the functon f(x) = 1 x2 x 2 4 on the axes gven below. There s a practce sheet of graph paper on the last sheet of ths exam that you may wsh to use to make a frst attempt at the graph. Soluton to Queston 3. (a) Set f (x) = 0 to get 6x = 0, so we have crtcal ponts when x = 0. We have sngular ponts when f (x) s undefned but f(x) = 0 s defned. Note that f (x) s undefned f (x 2 4) 2 = 0,.e x = ±2. But f(2), f( 2) are undefned, so there are no sngular ponts. (b) To fnd vertcal asymptotes, we look for 0 n the denomnator, so x = ±2. To fnd horzontal asymptotes, observe that the lm x ± f(x) = 1, so we have a horzontal asymptote y = 1. We only get oblque asymptotes f the degree of the numerator s equal to the degree of the denomnator plus 1, so there are no oblque asymptotes. (c) f s decreasng n (, 2) and ( 2, 0), and ncreasng n (0, 2) and (2, ). (d) f s concave up when 2 < x < 2, and concave down when x > 0 or x < 2.

7 Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 7 We don t have sngular ponts. The crtcal ponts are computed by settng T (x) = x 3 x = 0 5x = 3 x x > 0, x 2 = 9 x = 3. Here note that the rght hand sde of 5x = 3 x s postve, so x must be postve. We should compute T(3) = 7/5 + 5/3 = ( )/15 = 46/15 = 3 + 1/15. T (x) < 0 on x < 3. (you can plug n 0 nto T (x)) T (x) > 0 on x > 3. (you can take lmt x or plug n some number > 3, lke 10, and see that t s greater than 3 + 1/15.) Therefore x = 3 gves a local mnmum and there s no local maxmum. (c) x = 3 gves a mnmum of T(x) snce the functon s ncreasng on x > 3 and decreasng on x < 3. Therefore we should head to the pont 3 km west of the pont P. (d) Snce 3 + 1/15 s greater than 3, we can not make t to the bus.

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